The Spectral Radius of Graphs with No Intersecting Odd Cycles

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1.1 History and background In this section, we shall review the exact values of ex(n, F ) for some special graphs F , instead of the asymptotic estimation. A graph on 2k + 1 vertices consisting of k triangles which intersect in exactly one common vertex is called a k-fan (also known as the friendship graph) and denoted by Fk. Since χ(Fk) = 3, the Erd˝os-Stone-Simonovits theorem in (1) 2 2 implies that ex(n, Fk)= n /4+ o(n ). In 1995, Erd˝os et al. [15] proved the following exact result. Theorem 1.1. [15] For every k 1, and for every n 50k2, ≥ ≥ n2 k2 k, if k is odd, ex(n, F )= + − k 4 k2 3 k, if k is even. − 2 The extremal graphs of Theorem 1.1 are as follows. For odd k (where n 4k 1), ≥ − the extremal graph is uniquely constructed by taking a complete bipartite graph with color classes of size n and n and embedding two vertex disjoint copies of K in one side. For ⌈ 2 ⌉ ⌊ 2 ⌋ k even k (where now n 4k 3), the extremal graph in not unique, and each extremal graph ≥ − is constructed by taking a balanced complete bipartite graph and embedding a graph with 2k 1 vertices, k2 3 k edges with maximum degree k 1 in one side. − − 2 − Let Ck,q be the graph consisting of k cycles of length q which intersect exactly in one common vertex. Clearly, when we set q = 3, then Ck,3 is just the k-fan graph; see Theorem 1.1. When q is an odd integer, we can see that χ(Ck,q) = 3, the Erd˝os-Stone-Simonovits 2 2 theorem also implies that ex(n,Ck,q) = n /4 + o(n ). In 2016, Hou, Qiu and Liu [25] determined exactly the extremal number for C with k 1 and odd integer q 5. k,q ≥ ≥ Theorem 1.2. [25] For an integer k 1 and an odd integer q 5, there exists n (k,q) ≥ ≥ 0 such that for all n n (k,q), we have ≥ 0 n2 ex(n,C )= + (k 1)2. k,q 4 − Moreover, an extremal graph must be a Tur´an graph T2(n) with a Kk−1,k−1 embedding into one class.

2 We remark here that when q is even, then Ck,q is a bipartite graph where every vertex in one of its parts has degree at most 2. For such a sparse bipartite graph, a classical result 3/2 of Füredi [18] or Alon, Krivelevich and Sudakov [2] implies that ex(n,Ck,q) = O(n ). Recently, a breakthrough result of Conlon, Lee and Janzer [10, 11] shows that for even q 6 and k 1, we have ex(n,C ) = O(n3/2−δ) for some δ = δ(k,q) > 0. It is a ≥ ≥ k,q challenging problem to determine the value δ(k,q). For instance, the special case k = 1, this problem reduces to determine the extremal number for even cycle. Next, we shall introduce a unified extension of both Theorem 1.1 and Theorem 1.2. Let s, k be integers and let Hs,k be a graph consisting of s triangles and k cycles of odd lengths at least 5 which intersect in exactly one common vertex. The graph Hs,k is also known as the flower graph with s + k petals. We remark here that the k odd cycles can have different length. Clearly, when k = 0, then Hs,0 = Fs, the s-fan graph; see Theorem 1.1. In addition, when s = 0 and the lengths of odd cycles are all equal to q, then H0,k = Ck,q; see Theorem 1.2. In 2018, Hou, Qiu and Liu [26] and Yuan [42] independently determined the extremal number of H for s 0 and k 1. Let be the family of graphs with each member s,k ≥ ≥ Fn,s,k being a Turán graph T2(n) with a graph H embedded in one partite set, where

K , if (s, k) = (3, 1), H = s+k−1,s+k−1 6 (K3,3 or 3K3, if (s, k) = (3, 1), where 3K3 is the union of three disjoint triangles. Theorem 1.3. [26, 42] For two integers s 0, k 1, there exists n (s, k) such that for all ≥ ≥ 0 n n0(s, k), we have ≥ n2 ex(n,H )= + (s + k 1)2. s,k 4 − Moreover, the only extremal graphs for H are members of . s,k Fn,s,k 1.2 Spectral extremal problem Let G be a simple graph on n vertices. The adjacency matrix of G is deﬁned as A(G) = (aij)n×n with aij = 1 if two vertices vi and vj are adjacent in G, and aij = 0 otherwise. We say that G has eigenvalues λ1, . . . , λn if these values are eigenvalues of the adjacency matrix A(G). Let λ(G) be the maximum value in absolute among the eigenvalues of G, which is known as the spectral radius of graph G, that is, λ(G) = max λ : λ is an eigenvalue of G . {| | } By the Perron–Frobenius Theorem [24, p. 534], the spectral radius of a graph G is actually the largest eigenvalue of G since the adjacency matrix A(G) is nonnegative. The spectral radius of a graph sometimes can give some informations about the structure of graphs. For example, it is well-known [4, p. 34] that the average degree of G is at most λ(G), which is at most the maximum degree of G. In this paper we consider spectral analogues of Tur´an-type problems for graphs. That is, determining ex (n, F ) = max λ(G) : G = n, F * G . It is well-known that sp { | | } n ex(n, F ) ex (n, F ) (2) ≤ 2 sp

3 because of the fundamental inequality 2m λ(G). For most graphs, this study is again n ≤ fairly complete due in large part to a longstanding work of Nikiforov [36]. For example, he extended the classical theorem of Turán, by determining the maximum spectral radius of any Kr+1-free graph G on n vertices. The following problem regarding the adjacency spectral radius was proposed in [30]: What is the maximum spectral radius of a graph G on n vertices without a subgraph isomorphic to a given graph F ? Wilf [41] and Nikiforov [30] obtained spectral strengthening of Turán’s theorem when the forbidden substructure is the complete graph. Soon after, Nikiforov [31] showed that if G is a K -free graph on n vertices, then λ(G) λ(T (n)), r+1 ≤ r with equality if and only if G = Tr(n). Moreover, Nikiforov [31] (when n is odd), and Zhai and Wang [43] (when n is even) determined the maximum spectral radius of K2,2- free graphs. Furthermore, Nikiforov [33], Babai and Guiduli [3] independently obtained the spectral generalization of the K˝ovari-Sós-Turán theorem when the forbidden graph is the complete bipartite graph Ks,t. Finally, Nikiforov [34] characterized the spectral radius of graphs without paths and cycles of specified length. In addition, Fiedler and Nikiforov [16] obtained tight sufficient conditions for graphs to be Hamiltonian or traceable. For many other spectral analogues of results in extremal graph theory we refer the reader to the survey [36]. It is worth mentioning that a corresponding spectral extension [35] of the Erd˝os-Stone-Simonovits theorem states that 1 ex (n, F )= 1 + o(1) n. sp − χ(F ) 1 −

From this result, we know that exsp(n, Fk) = n/2 + o(n) where Fk is the k-fan graph. Recently, Cioab˘a, Feng, Tait and Zhang [7] generalized this bound by improving the error term o(n) to O(1), and obtained a spectral counterpart of Theorem 1.1. More precisely, they proved the following theorem.

Theorem 1.4. [7] Let G be a graph of order n that does not contain a copy of Fk where k 2. For suﬃciently large n, if G has the maximal spectral radius, then ≥ G Ex(n, F ). ∈ k

Recall that Hs,k is the graph consisting of s triangles and k cycles of odd lengths at least 5 which intersect in exactly one common vertex. Note that the k odd cycles can have diﬀerent length. In this paper, we shall prove the following theorem.

Theorem 1.5 (Main result). Let G be a graph of order n that does not contain a copy of H , where s 0 and k 1. For sufficiently large n, if G has the maximal spectral radius, s,k ≥ ≥ then G Ex(n,H ). ∈ s,k It is interesting that the spectral extremal example sometimes differs from the usual extremal example. For instance, Nikiforov [31], and Zhai and Wang [43] proved that the maximum spectral radius of a C4-free graph on n vertices is uniquely achieved by the friendship graph. This is very different from the usual extremal problem for the maximum number of edges in a C4-free graphs, since Füredi [21] showed that for n large enough with the form n = q2 + q + 1, the extremal number is uniquely attained by the polarity graph of

4 a projective plane. From Theorem 1.4 and Theorem 1.5, we know that graphs attaining the maximum spectral radius among all Fk-free (Hs,k-free) graphs also contain the maximum number of edges among all Fk-free (Hs,k-free) graphs. Our theorem is a spectral result of the Tur´an extremal problem for Hs,k, it can be viewed as an extension of Theorem 1.4, as well as a spectral analogue of Theorem 1.3. Our treatment strategy of the proof is mainly based on the stability method. To some extent, this paper could be regarded as a continuation and development of [7]. The heart of the proof and all key ideas lie in the proof of stability. We know that if we forbid the substructure Fk, then the neighbor of each vertex does not contain a matching of k edges. While we forbid the intersecting odd-length cycles, the neighbor of each vertex does not contain a long path, which can be viewed as a key observation in our extension. In addition, the embedding method of Hs,k is slightly diﬀerent from that of Fk, we need to prove the existence of a larger bipartite subgraph. We remark here that the spectral stability method is also used in a recent paper to deal with the extremal problem of odd-wheel graph [8].

2 Some Lemmas

In this section, we state some lemmas which are needed in our proof.

Lemma 2.1. [12] Let Pt denote the path on t vertices. If G is a Pt-free graph on n vertices, then e(G) (t−2)n , equality holds if and only if G is the disjoint union of copies of K . ≤ 2 t−1 Lemma 2.2. [7] If G has t triangles, then e(G) λ(G)2 3t . ≥ − λ(G) The next is the famous triangle removal lemma [37, 9, 17].

Lemma 2.3. [37] For every ε > 0, there exists δ(ε) > 0 such that every n-vertex graph with at most δ(ε) n3 triangles can be made triangle-free by removing at most εn2 edges. · Lemma 2.4 (Füredi [22]). Let G be a triangle-free graph on n vertices. If s > 0 and e(G)= e(T (n)) s, then there exists a bipartite subgraph H G such that e(H) e(G) s. 2 − ⊆ ≥ − Let G be a simple graph with matching number β(G) and maximum degree ∆(G). For given two integers β and ∆, define f(β, ∆) = max e(G) : β(G) β, ∆(G) ∆ . { ≤ ≤ } In 1976, Chvátal and Hanson [6] obtained the following result.

Lemma 2.5 (Chv´atal-Hanson [6]). For every two integers β 1 and ∆ 1, we have ≥ ≥ ∆ β f(β, ∆)=∆β + ∆β + β. 2 ∆/2 ≤ ⌈ ⌉ We will frequently use a special case proved by Abbott, Hanson and Sauer [1]:

k2 k, if k is odd, f(k 1, k 1) = − − − k2 3 k, if k is even. − 2 Furthermore, the extremal graphs attaining the equality case are exactly those we embedded into the Tur´an graph T2(n) to obtain the extremal Fk-free graph.

5 3 The Proof of Theorem 1.5

In the sequel, we always assume that G is a graph on n vertices containing no Hs,k as a subgraph and attaining the maximum spectral radius. The aim of this section is to prove that e(G) = ex(n,Hs,k) for n large enough. First of all, we note that G must be connected since adding an edge between different components will increase the spectral radius and also keep G being Hs,k-free. Let λ(G) be the spectral radius of G. By the Perron–Frobenius Theorem [24, p. 534], we know that λ1 has an eigenvector with all entries being positive, we denoted such an eigenvector by x. For a vertex v V (G), we will write x for the eigenvector entry of x corresponding to v. We ∈ v may normalize x so that it has maximum entry equal to 1, and let z be a vertex such that xz = 1. If there are multiple such vertices, we choose and fix z arbitrarily among them. In the sequel, we shall prove Theorem 1.5 iteratively, giving successively better lower bounds on both e(G) and the eigenvector entries of all of the other vertices, until finally we can show that e(G) = ex(n,Hs,k). The proof of Theorem 1.5 is outlined as follows.

2 We apply Lemma 2.2 to give a lower bound e(G) n O(n); see Lemma 3.1. Then ♠ ≥ 4 − we use the triangle removal lemma and Füredi’s stability result, and show that G has n n a very large bipartite subgraph on parts S, T with 2 o(n) S , T 2 + o(n). 2 − ≤ | | | | ≤ Moreover, we also have e(S, T ) n o(n2); see Lemma 3.2. ≥ 4 − We show that the number vertices that have Ω(n) neighbors on its side of the partition ♥ is bounded by o(n), and the number of vertices that have degree less than ( 1 O(1))n 2 − is bounded by O(1); see Lemma 3.3 and 3.4 respectively. Furthermore, we will prove that such vertices does not exist, and both G[S] and G[T ] are K1,s+k-free and Ms+k- free; see Lemma 3.6, 3.7 and 3.8. Based on the previous lemmas, we shall refine the structure of G, and improve the ♣ n2 n lower bound of e(G) to e(G) 4 O(1) and refine the bisection 2 O(1) S , T ≥2 − − ≤ | | | |≤ n + O(1) and also e(S, T ) n O(1); see Lemma 3.9. Moreover, we shall prove that 2 ≥ 4 − x = 1 o(1) for every u V (G); see Lemma 3.10. u − ∈ Once we know that all vertices have eigenvector entry close to 1, we can show that ♦ the bipartition is balanced; see Lemma 3.11, 3.12 and 3.13. This implies that G can be converted to a graph in Ex(n,Hs,k) by deleting few number of edges within S, T and adding few number edges between S and T . Invoking these facts, we finally show that e(G) = ex(n,Hs,k).

Let H be a Hs,k-free graph on n vertices with maximum number of edges. Since G is the graph maximizing the spectral radius over all Hs,k-free graphs, in view of Theorem 1.3, we can see by the Rayleigh quotient [24, p. 234] or [44, p. 267] that 1T A(H)1 2( n2/4 + (s + k 1)2) n λ(G) λ(H) = − > . (3) 1T 1 ≥ ≥ n 2 Lemma 3.1. Let c be the largest length of the cycles of Hs,k. Then n2 e(G) (s + k)cn. (4) ≥ 4 −

6 Proof. Since G is Hs,k-free, the neighborhood of any vertex does not contain P(s+k)c (a path on (s + k)c vertices) as a subgraph. Otherwise, G contains the join graph K P , 1 ∨ (s+k)c which contains a copy of Hs,k. Thus by Lemma 2.1, we can obtain the following upper bound for the number of triangles, (s + k)cn (s + k)c 3t = e(G[N(v)]) ex(n, P ) < = n2. ≤ (s+k)c 2 2 v∈XV (G) v∈XV (G) v∈XV (G) This gives t (s+k)c n2. From Lemma 2.2 and (3), we obtain ≤ 6 6t n2 e(G) λ2(G) (s + k)cn. (5) ≥ − n ≥ 4 − This completes the proof.

Lemma 3.2. Let ε be a fixed positive constant. There exists an N(ε, k) such that G has a partition V = S T which gives a maximum bipartite subgraph, and ∪ 1 e(S, T ) ε n2 ≥ 4 − for n N(ε, k). Furthermore ≥ 1 1 √ε n S , T + √ε n. (6) 2 − ≤ | | | |≤ 2 ε Proof. Let δ( 4 ) be the parameter chosen from the Triangle Removal Lemma 2.3. In the proof of Lemma 3.1, we know that t (s+k)c n3 δ( ε )n3 for n N = (s+k)c . By Lemma ≤ 6n ≤ 4 ≥ 6δ(ε/4) 2.3, there exists an N(ε, k) such that the graph G1 obtained from G by deleting at most ε n2 edges is K -free. For n N, the size of the graph G of order n satisfies 4 3 ≥ 1 ε n2 ε e(G ) e(G) n2 (s + k)cn n2. 1 ≥ − 4 ≥ 4 − − 4 Note that e(G ) e(T (n)) by the Mantel Theorem. We define s := e(T (n)) e(G ), then 1 ≤ 2 2 − 1 0 s (s + k)cn + ε n2. By Lemma 2.4, G contains a bipartite subgraph G such that ≤ ≤ 4 1 2 e(G ) e(G ) s. Hence, for n sufficiently large, we have 2 ≥ 1 − n2 ε 1 e(G ) e(G ) s (s + k)cn n2 ε n2. 2 ≥ 1 − ≥ 4 − − 2 ≥ 4 − Therefore, G has a partition V = S T which gives a maximum cut such that ∪ 1 e(S, T ) e(G ) ε n2. (7) ≥ 2 ≥ 4 − Furthermore, without loss of generality, we may assume that S T . If S < ( 1 √ε)n, | | ≤ | | | | 2 − then T = n S > ( 1 + √ε)n. So | | − | | 2 1 1 1 e(S, T ) S T < √ε n + √ε n = ε n2, ≤ | || | 2 − 2 4 − 7 which contradicts to Eq. (7). Therefore it follows that 1 1 √ε n S , T + √ε n. 2 − ≤ | | | |≤ 2 Hence the assertion (6) holds.

For a vertex v, let d (v) = N(v) S and d (v) = N(v) T . Next, we consider the S | ∩ | T | ∩ | set of vertices that have many neighbors which are not in the cut. Lemma 3.3. Let ε, δ be two suﬃciently small constants with ε < δ2/3. We denote

W := v S : d (v) δn v T : d (v) δn (8) { ∈ S ≥ } ∪ { ∈ T ≥ } For suﬃciently large n, we have 2δ 2(s + k 1)2 W n + − < δn. | |≤ 3 δn

2 Proof. Firstly, by Theorem 1.3, we know that e(G) ex(n,H ) n + (s + k 1)2. Note ≤ s,k ≤ 4 − that e(S, T ) 1 ε n2 by Lemma 3.2. Hence ≥ 4 − n2 e(S)+ e(T )= e(G) e(S, T ) + (s + k 1)2 1 ε n2 − ≤ 4 − − 4 − (9) = εn2 + (s + k 1)2. − On the other hand, if we denote by W = W S and W = W T , then we get 1 ∩ 2 ∩ 2e(S)= d (u) d (u) W δn, S ≥ S ≥ | 1| uX∈S uX∈W1 and similarly, we also have

2e(T )= d (u) d (u) W δn. T ≥ T ≥ | 2| uX∈T uX∈W2 So δn δn e(S)+ e(T ) ( W + W ) = W . (10) ≥ | 1| | 2| 2 2 | | Combining (9) and (10), we get δn W εn2 + (s + k 1)2, i.e., 2 | |≤ − 2εn2 + 2(s + k 1)2 W − . | |≤ δn Note that ε < δ2/3, we can get W < δn for suﬃciently large n. | | Lemma 3.4. Let k 2. Denote by ≥ 1 1 L := v V (G) : d(v) n . (11) ∈ ≤ 2 − 8c(s + k) Then L 16c2(s + k)2. | |≤

8 Proof. Suppose that L > 16c2(s + k)2. Then let L′ L with L′ = 16c2(s + k)2. Then it | | ⊆ | | follows that

e(G L′) e(G) d(v) − ≥ − ′ vX∈L n2 1 1 (s + k)cn 16c2(s + k)2 n ≥ 4 − − 2 − 8c(s + k) (n 16c2(s + k)2)2 > − + (s + k 1)2 4 − for suﬃciently large n, where the second inequality is by (5). Hence by Theorem 1.3, G L′ − contains Hs,k, which implies that G contains Hs,k. So the assertion holds. Now, we have proved that W = o(n) and L = O(1) by Lemmas 3.3 and 3.4, respec- | | | | tively. Next we will improve the bound on W and actually show that W is a subset of L, so W = O(1). To proceed, we ﬁrst need the following lemma which can be proved by | | induction or double counting.

Lemma 3.5. Let A , , A be p ﬁnite sets. Then 1 · · · p p A A A A (p 1) p A . | 1 ∩ 2 ∩···∩ p|≥ | i|− − |∪i=1 i| Xi=1 Lemma 3.6. Let W and L be sets of vertices deﬁned in (8) and (11). Then W L. ⊆ Proof. Suppose on the contrary that there exists a vertex u W and u / L. Let 0 ∈ 0 ∈ L = L S and L = L T . Without loss of generality, we may assume that u S, that 1 ∩ 2 ∩ 0 ∈ is, u W and u / L . Since S and T form a maximum bipartite subgraph, we have 0 ∈ 1 0 ∈ 1 d (u ) 1 d(u ). Indeed, otherwise, we can remove the vertex u into the part T , it will T 0 ≥ 2 0 increase strictly the number of edges between S and T . On the other hand, invoking the fact u L, we get d(u ) ( 1 1 )n. So 0 6∈ 0 ≥ 2 − 8c(s+k) 1 1 1 d (u ) d(u ) n. T 0 ≥ 2 0 ≥ 4 − 16c(s + k) Recall in Lemma 3.3 and 3.4 that

W < δn, L 16c2(s + k)2. | | | |≤ 1 Hence, for ﬁxed δ < 10(k+1)2 and suﬃciently large n, we have

1 S (W L) √ε n δn 16c2(s + k)2 (s + k)c. (12) | \ ∪ |≥ 2 − − − ≥ Claim. u is adjacent to at most s + k 1 vertices in S (W L). 0 − \ ∪ Suppose that u is adjacent to s+k vertices u , u , . . . , u in S (W L). Since u L, 0 1 2 s+k \ ∪ i 6∈ we have d(u ) ( 1 1 )n. On the other hand, we have d (u ) δn because u / W . i ≥ 2 − 8c(s+k) S i ≤ i ∈ So d (u )= d(u ) d (u ) ( 1 1 δ)n. In addition, we can choose other vertices T i i − S i ≥ 2 − 8c(s+k) −

9 u , . . . , u in the set S (W L), similarly, we also have d (u ) ( 1 1 δ)n s+k+1 (s+k)c \ ∪ S i ≥ 2 − 8c(s+k) − for each i [s + k + 1, (s + k)c]. By Lemma 3.5, we consider the common neighbors ∈ N (u ) N (u ) N (u ) T 0 ∩ T 1 ∩···∩ T (s+k)c (s+k)c (s+k)c N (u ) (s + k)c N (u ) ≥ | T i |− T i i=0 i=0 X [

dT (u0)+ dT (u1)+ + d T (u(s+k)c) ( s + k)c T ≥ · · · − | | 1 1 1 1 1 n + δ n (s + k)c (s + k)c + √ε n ≥ 4 − 16c(s + k) 2 − 8c(s + k) − · − 2 1 1 = (s + k)cδ (s + k)c√ε n> (s + k)c 8 − 16c(s + k) − − for suﬃciently large n, where the last inequality follows from the fact that δ and ε are small 1 δ2 enough, e.g., δ < 100c2(s+k)2 and ε< 3 . So there exist (s + k)c vertices v1, v2, . . . v(s+k)c in T such that the induced subgraph by two partitions u , . . . , u and v , . . . , v { 1 (s+k)c} { 1 (s+k)c} is complete bipartite. The subgraph of G formed by the vertex u0 together with such a complete bipartite graph can contain many disjoint odd-length cycles. For example, we can choose u0u1v1u0 to ﬁnd a copy of triangle, and we can choose u0u1v1us+k+1v2u0 to form a copy of pentagon and so on. Hence, it follows that G contains Hs,k, this is a contradiction. Therefore u is adjacent to at most s + k 1 vertices in S (W L). 0 − \ ∪ Hence, applying Lemmas 3.3 and 3.4 again, we have

d (u ) W + L + s + k 1 S 0 ≤ | | | | − 2δ 2(s + k 1)2 < n + − + 16c2(s + k)2 + s + k 1 3 δn − < δn for suﬃciently large n. This is a contradiction to the fact that u W . Similarly, there is 0 ∈ no vertex u such that u W and u / L . Hence W L. ∈ 2 ∈ 2 ⊆ Lemma 3.7. There exist independent sets I S and I T such that S ⊆ T ⊆ I S 20c2(s + k)2 and I T 20c2(s + k)2. | S| ≥ | |− | T | ≥ | |− Proof. Since S L is large enough by reviewing (12) in the proof of Lemma 3.6, we next prove \ that there exists a large complete bipartite subgraph between S and T . Let u1, . . . , u(s+k)c be (s + k)c vertices chosen arbitrarily from S L. Then u / L which implies that \ i ∈ 1 1 d(u ) n. i ≥ 2 − 8c(s + k) Note that W L by Lemma 3.6, so u / W , then d (u ) δn. Hence ⊆ i ∈ S i ≤ 1 1 d (u )= d(u ) d (u ) δ n. T i i − S i ≥ 2 − 8c(s + k) −

10 Furthermore, by Lemma 3.5, we have

(s+k)c (s+k)c (s+k)c N (u ) N (u ) ((s + k)c 1) N (u ) T i ≥ | T i |− − T i i=1 i=1 i=1 \ X [ 1 1 1 δ n (s + k) c ((s + k)c 1) + √ε n ≥ 2 − 8c(s + k) − · − − 2 3 = (s + k)cδ ((s + k)c 1)√ε n> (s + k)c 8 − − − for suﬃciently large n. Hence there exist (s + k)c vertices v1, v2, . . . , v(s+k)c such that the subgraph formed by two partitions u , . . . , u and v , . . . , v is a complete { 1 (s+k)c} { 1 (s+k)c} bipartite graph. Claim. G[S L] is both K -free and M -free. \ 1,s+k s+k Recall that G contains a large complete bipartite subgraph between S and T . If G[S L] \ contains a copy of K1,s+k centered at vertex u0 with leaves u1, u2, . . . , us+k, then by the discussion above, there exist ui and vj such that u0u1v1, u0u2v2, . . . , u0usvs form s triangles and v u v u v u u u forms an odd-length cycle and in fact we can ﬁnd all other 0 s+1 s+1 i j · · · x y 0 odd cycle similarly. Hence there is a H centered at u . Therefore, G[S L] is K -free. s,k 0 \ 1,s+k Now, we assume that u u , u u , . . . , u u is a matching of size s + k. Then { 1 2 3 4 2(s+k)−1 2(s+k)} u u v , . . . , u u v form s triangles, and v u u v u v u v forms an odd 1 2 1 2s−1 2s s s+1 2s+1 2s+2 i j · · · x y s+1 cycle and so on. So G[S L] is M -free. \ s+k Hence both the maximum degree and the maximum matching number of G[S L] are \ at most s + k 1, respectively. By Theorem 2.5, − e(G[S L]) f(s + k 1,s + k 1). \ ≤ − − The same argument gives

e(G[T L]) f(s + k 1,s + k 1). \ ≤ − − Since G[S L] has at most f(s + k 1,s + k 1) edges, then the subgraph obtained from \ − − G[S L] by deleting one vertex of each edge in G[S L] contains no edges, which is an \ \ independent set of G[S L]. By Lemma 3.4, there exists an independent set I S such \ S ⊆ that

I S L f(s + k 1,s + k 1) | S| ≥ | \ |− − − S 16c2(s + k)2 (s + k)2 S 20c2(s + k)2. ≥ | |− − ≥ | |− The same argument gives that there is an independent set I T with T ⊆ I T 20c2(s + k)2. | T | ≥ | |− This completes the proof.

In Lemma 3.7, we have showed that there are two large independent set with ( 1 o(1))n 2 − vertices both in the sets S and T . Invoking this fact, we next shall prove that L is actually an empty set.

11 Lemma 3.8. L is empty, and both G[S] and G[T ] are K1,s+k-free and Ms+k-free.

Proof. Recall that Ax = λ1x and z is deﬁned as a vertex with maximum eigenvector entry and satisﬁes xz = 1. So we have n d(z) x = λ x = λ . ≥ w 1 z 1 ≥ 2 w∼z X Hence z / L. Without loss of generality, we may assume that z S. Since the maximum ∈ ∈ degree in the induced subgraph G[S L] is at most s + k 1 (containing no K ), from \ − 1,s+k Lemma 3.4, we have L 16c2(s + k)2 and | |≤ d (z)= d (z)+ d (z) 16c2(s + k)2 + s + k 1 20c2(s + k)2. S S∩L S\L ≤ − ≤ Therefore, by Lemma 3.7, we have

λ1 = λ1xz = xv = xv + xv v∼z X v∼Xz,v∈S v∼Xz,v∈T = xv + xv + xv v∼Xz,v∈S v∼Xz,v∈IT v∼z,vX∈T \IT d (z)+ x + 1 ≤ S v vX∈IT v∈XT \IT 20c2(s + k)2 + x + T I ≤ v | | − | T | vX∈IT x + 40c2(s + k)2. ≤ v vX∈IT Combining (3), we can get n x 40c2(s + k)2. (13) v ≥ 2 − vX∈IT Next we are going to prove L = ∅. By way of contradiction, assume that there is a vertex v L, so d (v) ( 1 1 )n. ∈ G ≤ 2 − 8c(s+k) Consider the graph G+ with vertex set V (G) and edge set E(G+) = E(G v ) vw : 1 \ { } ∪ { w IT . Roughly speaking, in this process, we have deleted ( 2 O(1))n edges and added 1 ∈ } − ( 2 o(1))n edges. Note that adding a vertex incident with vertices in IT does not create − + any triangles and odd cycles, and so G is Hs,k-free. Note that x is a vector such that T T + x A(G)x + x A(G )x λ(G)= T , and the Rayleigh theorem implies λ(G ) T . Furthermore, x x ≥ x x

xT (A(G+) A(G)) x 2x λ(G+) λ(G) − = v x x − ≥ xTx xT x  w − u wX∈IT uv∈XE(G) (13) 2x n   v 40c2(s + k)2 d (v) ≥ xT x 2 − − G 2x n 1 1 v 40c2(s + k)2 ( )n ≥ xT x 2 − − 2 − 8c(s + k)

12 2x n = v 40c2(s + k)2 > 0, xT x 8c(s + k) − where the last inequality holds for n large enough. This contradicts G has the largest spectral radius over all Hs,k-free graphs, so L must be empty. Furthermore, the claim in the proof of Lemma 3.7 implies that both G[S] and G[T ] are K1,s+k-free and Ms+k-free.

Let G be a Hs,k-free graph on n vertices with maximum spectral radius. In the previous 2 n2 lemmas, we have proved that G contains at most O(n ) triangles and has 4 O(n) edges. − n In addition, G contains a bipartite subgraph with parts S and T such that 2 o(n) n − ≤ S , T 2 + o(n). Next we shall reﬁne the structure of G. We shall show that the number | | | |≤ 2 of triangles in G is at most O(n) and the number of edges in G is at least n O(1), and 4 − the two vertex parts S, T satisﬁes n O(1) S , T n + O(1). More precisely, we state 2 − ≤ | | | |≤ 2 these results as in the following lemma.

Lemma 3.9. For n and k deﬁned as before, we have

n2 e(G) 12(s + k)2, ≥ 4 − n2 e(S, T ) 14(s + k)2. ≥ 4 − n n 4(s + k) S , T + 4(s + k), 2 − ≤ | | | |≤ 2 and n n 14(s + k)2 δ(G) λ ∆(G) + 5(s + k). 2 − ≤ ≤ 1 ≤ ≤ 2

Proof. From Lemma 3.8, both G[S] and G[T ] are K1,s+k-free and Ms+k-free. By lemma 2.5, so we have e(S)+ e(T ) 2f(s + k 1,s + k 1) < 2(s + k)2. This means that the ≤ − − number of triangles in G is bounded above by 2(s+k)2n since any triangle contains an edge of E(S) E(T ). By Lemma 2.2, we have ∪ 6t n2 e(G) λ2 12(s + k)2. ≥ 1 − n ≥ 4 − Since e(S)+ e(T ) 2(s + k)2, then we have ≤ n2 e(S, T )= e(G) e(S) e(T ) 14(s + k)2. − − ≥ 4 − Suppose that S n 4(s + k), then T = n S n + 4(s + k). Hence | |≤ 2 − | | − | |≥ 2 n n n2 e(S, T ) S T 4(s + k) + 4(s + k) = 16(s + k)2, ≤ | || |≤ 2 − 2 4 − 2 which contradicts to e(G) n 14(s + k)2. So we have ≥ 4 − n n 4(s + k) S , T + 4(s + k). 2 − ≤ | | | |≤ 2

13 Moreover, by Lemma 3.8, the maximum degree of G[S] and G[L] is at most s + k 1, which − yields n n ∆(G) ( + 4(s + k)) + (s + k 1) < + 5(s + k). ≤ 2 − 2 So n λ ∆(G) < + 5(s + k). 1 ≤ 2 Furthermore, we claim that the minimum degree of G is at least n 14(s + k)2. Otherwise, 2 − removing a vertex v of minimum degree d(v), we have

e(G v) = e(G) d(v) − − n2 n 12(s + k)2 14(s + k)2 ≥ 4 − − 2 − n2 n = + 2(s + k)2 4 − 2 (n 1)2 > − + (s + k 1)2, 4 − which implies the induced subgraph G v contains a copy of H by Theorem 1.3. − s,k 2 Lemma 3.10. For all u V (G), we have that x 1 120(s+k) . ∈ u ≥ − n Proof. Without loss of generality, we may assume that z S. We consider the following ∈ two cases. Step 1. We ﬁrst consider the case u S. Since G[S] is K -free, then d (u) ∈ 1,s+k S ≤ s + k 1. By Lemma 3.9, we have −

NT (u) = dT (u)= d(u) dS(u) δ(G) dS(u) | | n − ≥ − 14(s + k)2 (s + k 1) ≥ 2 − − − n 15(s + k)2. ≥ 2 − Similarly, we also have N (z) n 15(s + k)2. Then | T |≥ 2 −

NT (u) NT (z) = NT (u) + NT (z) NT (u) NT (z) | ∩ | | n | | | − | n ∪ | 2 15(s + k)2 + 4(s + k) ≥ 2 − − 2 n 34(s + k)2. ≥ 2 − Note that d (z) T . By Lemma 3.9 again, we can get T ≤ | | n n d (z) N (u) N (z) + 4(s + k) ( 34(s + k)2) 38(s + k)2. T − | T ∩ T |≤ 2 − 2 − ≤ Hence, we have

λ x λ x = x x 1 u − 1 z v − z v∼u v∼z X X

14 = x + x x x v v − v − v v∼u,vX∈T,v6∼z v∼Xu,v∈S v∼z,vX∈T,v6∼u v∼Xz,v∈S x x ≥ − v − v v∼z,vX∈T,v6∼u v∼Xz,v∈S 1 1 ≥ − − v∼z,vX∈T,v6∼u v∼Xz,v∈S d (z) N (u) N (z) d (z) ≥ − T − | T ∩ T | − S 38( s + k)2 (s + k)2 ≥ − − = 39(s + k)2. − Note that x = 1. Therefore, for any u S, we have z ∈ 39(s + k)2 78(s + k)2 78(s + k)2 xu 1 > 1 n = 1 . (14) ≥ − λ1 − 2 − n Step 2. Now we consider the case u T . By (14), we get ∈ 78(s + k)2 λ x = x x 1 d (u). 1 u v ≥ v ≥ − n S v∼u X v∼Xu,v∈S By Lemma 3.9, we can see that d(u) δ(G) n 14(s+k)2. Recall that G[T ] is K -free, ≥ ≥ 2 − 1,s+k so we have d (u) s + k 1. Then T ≤ − n d (u)= d(u) d (u) 15(s + k)2. S − T ≥ 2 − Hence

2 2 78(s+k) 78(s+k) n 2 (1 n )dS (u) (1 n )( 2 15(s + k) ) xu − − n − ≥ λ1 ≥ 2 + 5(s + k) 4 n 2 1170(s+k) 2 54(s + k) + n = − n 2 + 5(s + k) 120(s + k)2 > 1 . − n From the above two cases, the result follows.

Using this reﬁned bound on the eigenvector entries, we will show that the partition V = S T is balanced (Lemma 3.13). First of all, we ﬁx some notation for convenience. ∪ Let B = Ks,t be the complete bipartite graph with partite sets S and T , and let G1 = G[S] G[T ] and G be the graph on V (G) with the missing edges between S and T , that ∪ 2 is, E(G )= E(B) E(G). Note that e(G)= e(G )+ e(B) e(G ). 2 \ 1 − 2 From Lemma 3.8, we know that both G[S] and G[T ] are K1,s+k-free and Ms+k-free, then e(G ) = e(S)+ e(T ) 2f(s + k 1,s + k 1) 2(s + k)2. Next we shall give an 1 ≤ − − ≤ improvement in the sense that e(G2) is closed to zero.

15 Lemma 3.11. Let G1, G2 and B be graphs deﬁned in above. Then

e(G ) e(G ) (s + k 1)2. 1 − 2 ≤ − Proof. Without loss of generality, we may assume that T S and denote by | | ≥ | | S′ := v S : N(v) T , { ∈ ⊆ } T ′ := v T : N(v) S . { ∈ ⊆ } Since e(G[S]) f(s+k 1,s+k 1) (s+k)2 by Lemma 3.8, there exist at most 2(s+k)2 ≤ − − ≤ vertices in S having a neighbor in S. Hence

S′ S 2(s + k)2. | | ≥ | |− Similarly, T ′ T 2(s + k)2. | | ≥ | |− Let C T ′ be a set having T S vertices, which is well-deﬁned since we can see from ⊆ | | − | | Lemma 3.9 that T S 8(s + k) and T ′ T 2(s + k)2 n 4(s + k) 2(s + k)2 > | | − | |≤ | | ≥ | |− ≥ 2 − − 8(s + k). Then G C is a graph on 2 S vertices such that \ | | (2 S )2 e(G) e(C,S)= e(G C) ex(2 S ,H ) | | + (s + k 1)2. − \ ≤ | | s,k ≤ 4 − Hence e(G) S 2 + C S + (s + k 1)2 = S T + (s + k 1)2. ≤ | | | || | − | || | − Note that e(G ) e(G )= e(G) e(B). This completes the proof. 1 − 2 − Lemma 3.12. 2 n2 7200(s + k)4 S T . n 4 − | || |≤ n(n 240(s + k)2) p − Proof. By Lemma 3.10 we have,

120(s + k)2 2 240(s + k)2 xT x n 1 >n 1 = n 240(s + k)2, (15) ≥ − n − n − and that λ(B)= S T . By Lemma 3.9, we know that e(G ) 2(s + k)2, we obtain | || | 1 ≤ p n2 n2 e(S, T )= e(G) e(G ) 12(s + k)2 2(s + k)2 = 14(s + k)2, − 1 ≥ 4 − − 4 − which implies that

n2 e(G )= e(B) e(S, T ) S T 14(s + k)2 14(s + k)2. 2 − ≤ | || |− 4 − ≤ Applying Lemma 3.10 again, we can obtain

120(s + k)2 2 xT A(G )x = 2 x x 2e(G ) 1 2 u v ≥ 2 − n uv∈XE(G2)

16 240(s + k)2 2e(G ) 1 . ≥ 2 − n Combining this result together with (3) and Lemma 3.11, we can get

2 n2 2(s + k 1)2 (3) xT (A(B)+ A(G ) A(G ))x + − λ(G)= 1 − 2 n 4 n ≤ xT x xT A(B)x xT A(G )x xT A(G )x = + 1 2 xT x xT x − xT x 240(s+k)2 2e(G ) 2e(G2)(1 ) λ(B)+ 1 − n ≤ xT x − xT x 240(s+k)2 2(e(G ) e(G )) 2e(G2) λ(B)+ 1 − 2 + n ≤ xT x xT x 240(s+k)2 Lemma 3.11 2(s + k 1)2 2 14(s + k)2 S T + − + · n . ≤ | || | xT x xT x p Then we have

2 2 n2 1 1 28(s + k)2 240(s+k) S T 2(s + k 1)2 + n n 4 − | || |≤ − xT x − n xT x p (15) 1 1 6720(s + k)4 2(s + k)2 + ≤ n 240(s + k)2 − n n(n 240(s + k)2) − − 480(s + k)4 6720(s + k)4 = + n(n 240(s + k)2) n(n 240(s + k)2) − − 7200(s + k)4 = . n(n 240(s + k)2) − This completes the proof.

Lemma 3.13. The sets S and T have sizes as equal as possible. That is

S T 1. | | − | | ≤ Proof. We assume on the contrary that T S + 2. We consider two cases. | | ≥ | | Case 1: n is even. Since S + T = n, we have | | | | 2 n2 n n n S T 1 + 1 n 4 − | || | ≥ 2 − 2 − 2 r p n n2 1 1 = 1= > . 2 2 − r 4 − n + n 1 n 2 4 − q So by Lemma 3.12, we have

1 2 n2 7200(s + k)4 < S T . n n 4 − | || |≤ n(n 240(s + k)2) p −

17 This is a contradiction for sufficiently large n. Case 2: n is odd. Since S + T = n, we have | | | | 2 n2 n2 1 n 3 n + 3 S T − − n 4 − | || | ≥ 2n − 2 2 s p 1 1 (n 1 )2 (n2 9) = n n2 9 = − n − − 2 − n − − 2(n 1 + √n2 9) − n − 1p 7+ 2 1 = n . 2(n 1 + √n2 9) ≥ n − n − So by Lemma 3.12 again, we get 1 2 n2 7200k4 < S T . n n 4 − | || |≤ n(n 240k2) p − This is a contradiction for sufficiently large n. Therefore for n large enough we must have that S T 1. || | − | || ≤ Recall that G is an Hs,k-free graph with the maximum spectral radius. Finally, we will show that e(G) = ex(n,Hs,k). In other words, G also attains the maximum number of edges among all Hs,k-free graphs. Proof of Theorem 1.5. By way of contradiction, we may assume that e(G) ≤ ex(n,H ) 1. By Lemma 3.13, we know that S T 1. Let H be an H -free s,k − | | − | | ≤ s,k graph with ex(n,H ) edges on the same vertex set as G such that the crossing edges be- s,k tween S and T span a complete bipartite graph in H, this is possible because every graph in Ex(n,H ) has a maximum cut of size n2/4 by Theorem 1.3. Let E and E be sets of s,k ⌊ ⌋ + − edges such that E(G) E E = E(H), where E = E(H) E(G) and E = E(G) E(H). ∪ + \ − + \ − \ Note that e(G)+ E E = e(H), which together with e(H) e(G) + 1 implies that | +| − | −| ≥ E E + 1. | +| ≥ | −| 2 Furthermore, we have that E− e(G[S]) + e(G[T ]) < 2(s + k) . By Lemma 3.9, we have 2 | |≤ that E = n e(S, T ) 14(s + k)2. Now, by Lemma 3.10, we have that | +| ⌊ 4 ⌋− ≤ xT A(H)x 2 2 λ(H) = λ(G)+ x x x x ≥ xT x xT x i j − xT x i j ijX∈E+ ijX∈E− Lemma 3.10 2 120(s + k)2 2 λ(G)+ E 1 E ≥ xT x | +| − n − | −| 2 240(s + k)2 (120(s + k)2)2 λ(G)+ E E E + E ≥ xT x | +| − | −|− n | +| n2 | +| 2 240(s + k)2 (120(s + k)2)2 λ(G)+ 1 E + E ≥ xT x − n | +| n2 | +| > λ(G) for sufficiently large n, where the last inequality follows by E < 14(s + k)2. Therefore | +| we have that for n large enough, λ(H) > λ(G), a contradiction. Hence e(G) = e(H). By Theorem 1.3, we know that G Ex(n,H ). The proof of Theorem 1.5 is complete. ∈ s,k

18 4 Concluding remarks

To avoid unnecessary calculations, we did not attempt to get the best bound on the order of graphs in the proof. Our proof used the Triangle Removal Lemma, which means that the condition “sufficiently large n” is needed in our proof. It is interesting to determine how large n needs to be for our result. Recently, Cioab˘a, Desai and Tait [8] investigated the largest spectral radius of an n- vertex graph that does not contain the odd-wheel graph W2k+1, which is the graph obtained by joining a vertex to a cycle of length 2k. Moreover, they raised the following more general conjecture. Conjecture 4.1. Let F be any graph such that the graphs in Ex(n, F ) are Turán graphs adding O(1) edges. Then for sufficiently large n, a graph attaining the maximum spectral radius among all F -free graphs is a member of Ex(n, F ). We say that F is edge-color-critical if there exists an edge e of F such that χ(F e) < − χ(F ). Let F be an edge-color-critical graph with χ(F )= r + 1. By a result of Simonovits [38] and a result of Nikiforov [35], we know that Ex(n, F ) = EX (n, F ) = T (n) for sp { r } sufficiently large n, this shows that Conjecture 4.1 is true for all edge-color-critical graphs. As we mentioned before, Theorem 1.4 says that Conjecture 4.1 holds for the k-fan graph Fk. In addition, our main result (Theorem 1.5) tells us that Conjecture 4.1 also holds for the flower graph Hs,k. Note that both Fk and Hs,k are not edge-color-critical. Let S be the graph consisting of a clique on k vertices and an independent set on n k n,k − vertices in which each vertex of the clique is adjacent to each vertex of the independent set. Clearly, we can see that Sn,k does not contain Fk as a subgraph. Recently, Zhao, Huang and Guo [45] proved that Sn,k is the unique graph attaining the maximum signless Laplacian spectral radius among all graphs of order n containing no F for n 3k2 k 2. So it k ≥ − − is a natural question to consider the maximum signless Laplacian spectral radius among all graphs containing no Ck,q, the graph defined as k cycles of odd-length q intersecting in a common vertex. We write q(G) for the signless Laplacian spectral radius, i.e., the largest eigenvalue of the signless Laplacian matrix Q(G) = D(G)+ A(G), where D(G) = diag(d1,...,dn) is the degree diagonal matrix and A(G) is the adjacency matrix. We end with the following conjecture (Clearly, when t = 1, our conjecture reduces to the result of Zhao et al. [45]). Conjecture 4.2. For integers k 2,t 1 and q = 2t + 1, there exists an integer n (k,t) ≥ ≥ 0 such that if n n (k,t) and G is a C -free graph on n vertices, then ≥ 0 k,q q(G) q(S ), ≤ n,kt equality holds if and only if G = Sn,kt.

Another interesting problem on this topic is to determine the Tur´an number of Ck,q for even q. More general, it is challenging to determine the Tur´an number of Hs,k where the cycles have even lengths.

Acknowledgements The ﬁrst author would like to express his sincere thanks to Prof. Lihua Feng and Lu Lu for many illuminating discussions. This work was supported by NSFC (Grant No. 11931002).

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